What is the integral from 0 to 5 of (4t)/((t-6)^2) ?

The integral from 0 to 5 of (4t)/((t-6)^2) is 4(-ln(6)+5)

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integral from 0 to 5 of (4t)/((t-6)^2)

\int_{0}^{5} \frac{4 t}{( t - 6 ) ^{2}} d t

4 (- ln (6) + 5)

Decimal

12.83296 \dots

Solution steps

\int_{0}^{5} \frac{4 t}{( t - 6 ) ^{2}} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{0}^{5} \frac{t}{( t - 6 ) ^{2}} d t

Apply u-substitution

: \int_{- 6}^{- 1} \frac{u + 6}{u ^{2}} d u

= 4 \cdot \int_{- 6}^{- 1} \frac{u + 6}{u ^{2}} d u

Expand

\frac{u + 6}{u ^{2}} : \frac{1}{u} + \frac{6}{u ^{2}}

= 4 \cdot \int_{- 6}^{- 1} \frac{1}{u} + \frac{6}{u ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int_{- 6}^{- 1} \frac{1}{u} d u + \int_{- 6}^{- 1} \frac{6}{u ^{2}} d u)

\int_{- 6}^{- 1} \frac{1}{u} d u = - ln (6)

\int_{- 6}^{- 1} \frac{6}{u ^{2}} d u = 5

= 4 (- ln (6) + 5)

\int \frac{( 4 u ^{2} - 8 )}{2 u ^{2} + 1} d u

\frac{d}{d x} (\frac{x ^{2}}{4 - x ^{2}})

- \frac{24}{x ^{4}}

\int p (p + 3)^{6} d p

x \to 4 lim (x - 4)

What is the integral from 0 to 5 of (4t)/((t-6)^2) ?
The integral from 0 to 5 of (4t)/((t-6)^2) is 4(-ln(6)+5)